7
$\begingroup$

Let $U$ and $U'$ be $S^4 - x_N$ and $S^4 - x_S$, respectively, where $x_N$ is the North pole and $x_S$ is the South pole. The usual stereographic projection maps $U$ into $R^4$ and $U'$ into $R^4$. If we identify $R^4$ with the quaternions, then we can write the coordinates on $U$ as a quaternion $z$ and the coordinates on $U'$ as another quaternion $z'$. Then, the change of coordinates between these two charts take the remarkably simple form $z'=1/z$. This is well known, and can be found in many references.

My question regards a different choice of coordinates on $S^4$, which I found on an old paper, that unfortunately skips the details on how its constructed. The two patches $U$ and $U'$ are defined in the same way, but they are mapped to the disk $\{x \in R^4 | |x|^2 <1 \}$, rather than the full $R^4$. Such a region is again identified with quaternions via the usual isomorphism; let's say that the quaternion $z$ corresponds to the coordinates on $U$ and $z'$ corresponds to those on $U'$, as before. The transition function reads: $z'= \frac{1 - ||z||}{||z||}z$, where $||z||$ is the norm of $z$.

For reference, this question originated when reading the paper "Homeomorphy classification of total spaces of sphere bundles over sphere" by Tamura, page 31 (just before section 2).

I tried a combination of stereographic coordinates with some conformal mapping of $R^4$ into the unit disk, but I could not obtain the transition function above. I have a feeling that it should be something simple, but I can't think of something that matches the properties that I described. Any suggestions on how these coordinates might be defined?

Thanks in advance!

$\endgroup$
24
  • 1
    $\begingroup$ @Semiclassical sure, it is "Homeomorphy classification of total spaces of sphere bundles over sphere" by Tamura, page 31 (just before section 2). $\endgroup$
    – User175a23
    Apr 2 at 19:43
  • 1
    $\begingroup$ Would it be ok with you to simply understand an easy diffeomorphism between an open ball in $\mathbb R^n$ and the whole $\mathbb R^n$? E.g., $x\to x/(1-|x|)$ or similar? $\endgroup$ Apr 2 at 19:45
  • 1
    $\begingroup$ @PaulFrost, perhaps I'm misreading the question. Is the desired transition function not $$z' = \frac{1-|z|}{|z|}z?$$ That's radial scaling. $\endgroup$
    – Deane
    Apr 3 at 12:43
  • 1
    $\begingroup$ @User175a23 If you allow different transformations $\mathbb R^4 \to B^4$ to adjust the stereographic projections, then you get probably quite a number of solutions. $\endgroup$
    – Paul Frost
    Apr 3 at 17:30
  • 1
    $\begingroup$ @PaulFrost, indeed. You can choose the parameterization of the vertical angle to be any monotone map from $[0,1]$ to $[0,\pi]$ to define the map to $U$. That then uniquely determines the map to $U'$. $\endgroup$
    – Deane
    Apr 3 at 17:43

1 Answer 1

3
$\begingroup$

You claimed that the change of coordinates between the two stereographic projection charts take the remarkably simple form $z′=1/z$. This is false with the standard interpretation of stereographic projection. It takes the form $$z' = \frac{z}{\lvert z \rvert^2} \tag{1}$$ which can be rewritten as $$z' = \frac{1}{\overline z} \tag{2}$$ using quaternion multiplication. Concerning $(1)$ see my answer to Two equal smooth structures on $S^n$ for the more general case of the stereographic projections $$\sigma_N : S^n \setminus \{x_N\} \to \mathbb R^n, \sigma_N(x,x_{n+1}) = \frac{x}{1 - x_{n+1}} ,$$ $$\sigma_S : S^n \setminus \{x_S\} \to \mathbb R^n,\sigma_S(x,x_{n+1}) = \frac{x}{1 + x_{n+1}} .$$ Here points $\xi \in \mathbb R^{n+1}$ are written in the form $\xi = (x,x_{n+1}) \in \mathbb R^n \times \mathbb R$. See also Wikipedia.

With $z = \sigma_N(\xi)$ and $z' = \sigma_S(\xi)$ we get $$z' = \frac{z}{\lVert z \rVert^2}\tag{1'}$$

Here we used the Euclidean norm $\lVert z \rVert$ which agrees with the abolute value $\lvert z \rvert$ of quaternions for $n = 4$. $(1')$ is due to the relation $\sigma_S(x,x_{n+1}) = \sigma_N(x,-x_{n+1})$.

For $n = 4$ we would get your equation $z' = \frac{1}{z}$ if we define $\sigma_S(x,x_{n+1}) = \frac{1}{1 + x_{n+1}} \overline x$ with the conjugate quaternion $\overline x$. See Constructing the Riemann Sphere for the complex case ($n = 2$). There may be some use for this alternative definition of $\sigma_S$; in the complex case we get a holomorphic coordinate change.

Tamura certainly uses the standard stereographic projections and composes them with $$\phi : \mathbb R^n \to B^n, \phi(y) = \frac{y}{1 + \lVert y \rVert} .$$ Its inverse is given by $\phi^{-1}(z) = \frac{z}{1 - \lVert z \rVert}$.

So let us consider the charts $\tau_N = \phi \circ \sigma_S$ and $\tau_S = \phi \circ \sigma_S$. We get $$\tau_S \circ \tau_N^{-1} = \phi \circ \sigma_S \circ \sigma_N^{-1} \circ \phi^{-1}$$ and therefore $$z' = (\tau_S \circ \tau_N^{-1})(z) = \phi(\frac{\phi^{-1}(z)}{\lVert \phi^{-1}(z) \rVert^2}) =\phi(\frac{(1 - \lVert z \rVert) z}{\lVert z \rVert^2}) = \frac{1 -\lVert z \rVert}{\lVert z \rVert}z .$$

With the alternative definition $\sigma_S(x,x_{n+1}) = \frac{1}{1 + x_{n+1}} \overline x$ we get $$z' = \frac{1 -\lVert z \rVert}{\lVert z \rVert}\overline z .$$

$\endgroup$
4
  • $\begingroup$ Thank you for your answer! I understand and agree with your concerns: with the standard interpretation (i.e. the perspective projection of the points on the sphere except the pole onto the plane which lies at the equator) gives $z'=\frac{1}{\bar{z}}$. Many authors add minus signs (not only in the quaternionic version, but also in the complex case for $S^2$), to achieve the "nicer" transition functions $z'=\frac{1}{z}$. I was referring to these initially. This was not the main point of the question, clearly, but I realise that it has been a point of confusion for some, so I edited the question. $\endgroup$
    – User175a23
    Apr 4 at 9:52
  • $\begingroup$ @User175a23 Perhaps you should add a remark to your question that you initially worked with versions of stereographic projection producing $z' = 1/z$. In the present form it is not clear why you asked. $\endgroup$
    – Paul Frost
    Apr 4 at 10:05
  • $\begingroup$ @User175a23 Editing questions to correct typos or to add additional material is commendable. But your last edit completely changed the game, and you also provided the answer in your question. I think one should not do this. In my comment to your question I was confused about Deane's claim ""unlike stereographic projection ...". Paul Frost's answer explained it to me. Your "new" question will probably cause new confusion. Readers will be confused because nothing is open now. $\endgroup$ Apr 4 at 10:19
  • $\begingroup$ That's right, new readers would have found it confusing indeed. I edited the question, bringing it back to its original form and deleting the edit part. I added it to give credits to the answers given in the comments, but I appreciate that, since the chronological order is not clear, the logical order is also lost. Sorry about that, I am not an experienced user of this platform, thanks for pointing it out! $\endgroup$
    – User175a23
    Apr 4 at 11:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .